The materials to prepare students for practical lessons of inorganic chemistry

LESSON13

Theme. Experimental design of redox-reaction. Review of general chemistry

Plan

Experimental study of redox reactions.

Experimental study of complex compounds.

 

Types of Reactions: Redox and Non-Redox

 

Introduction.

What is a redox reaction?

The formal name for a redox reaction is “oxidation reduction reaction,” and you can see that “redox” is just shorthand for the words reduction and oxidation. Thus, in a redox reaction, two things happen. You guessed it — oxidation and reduction. These two have to happen together. You cannot have an oxidation reaction without a corresponding reduction reaction. It’s a bit like the idea behind a blood transfusion or an organ transplant. You cannot have a recipient unless you have a donor, and it does not make any sense to be a donor unless there is a recipient.

You may have noticed that “oxidation” starts with the same prefix as oxygen, suggesting that oxygen may be somehow involved in this process. Indeed, one definition of oxidation states that:

Oxidation is the gaining of bonds to oxygen

Organic fuel substances (such as wood, coal or gas) are examples of compounds that can be oxidized. In the process of burning, the carbon in these substances becomes bonded with oxygen, while some of the oxygen used to “burn” the fuel bonds to the hydrogen atoms from the fuel. Thus, one of the definitions of reduction states that:

Reduction is the gaining of hydrogen

Therefore, combustion reactions are good examples of redox reactions where one molecule gains oxygen (is oxidized) and one molecule gains hydrogen (is reduced).

For example, let’s look at what happens when gasoline in your car is burned as you drive around town (heptane is a common hydrocarbon component of gasoline):

Note how the carbon atoms in heptane are oxidized (because the carbon atoms in heptane become bonded with oxygen atoms), while the oxygen is reduced (becomes bonded with hydrogen atoms).

These definitions of oxidation and reduction are useful. However, more general definitions of oxidation and reduction involve the movement of electrons between the compounds involved in the redox reaction. In the most broad definition of redox reactions:

Oxidation is the loss of electrons. Reduction is the gain of electrons

Because electrons are negatively charged, an increase in electrons means a decrease in overall charge (the compound becomes more negatively charged). On the other hand, an atom that is oxidized has given up some of those negatively charged electrons, which will increase its overall charge (the compound becomes more positively charged). Notice that these definitions do not involve oxygen and hydrogen. Thus, redox reactions can occur with compounds that do not contain oxygen or hydrogen atoms.

An everyday example of a redox reaction that we are all familiar with is the process of rusting. Rust is the flaky brown substance that forms on iron objects left exposed to the elements for too long, especially if the objects get wet. Rust doesn’t just form on the iron object, the iron actually turns into rust (rust is actually a form of oxidized iron).

We all intuitively know the chemical reaction of rusting:

Let’s examine the oxidation and reduction reactions that are involved in the redox reaction of rusting. First, iron, in the presence of moisture (H₂O) will lose electrons, becoming a positively charged ion in water:

Oxidation reaction: Iron is oxidized (loses e-)

Fe_(s) = Fe₂⁺_(aq) + 2 e^–

Those electrons are then used to reduce the oxygen dissolved in the water (remember that water will have some dissolved oxygen and also H⁺ and OH^– ions):

Reduction reaction: Oxygen is reduced (gains e-)

O₂ + 4e^– + 4 H⁺ = 2 H₂O

Those Fe₂⁺ ions react with the OH^– ions in water to produce iron hydroxide, which will dry in several steps to produce rust:

Fe₂⁺ + 2OH^– = Fe(OH)₂ = =  Fe₂O₃ (rust)

Notice that the redox reactioeeds water, which explains why a moist environment speeds up the rusting process. Rusting can occur in dry climates, but it tends to happen much more slowly due to the relatively low humidity in the air.

Because it is difficult to remember the definitions of oxidation and reduction that involve the movement of electrons, a common mnemonic that can help you keep these rules in mind is:

“OIL RIG”

This is an especially appropriate mnemonic for remembering redox reactions because the burning of oil (a hydrocarbon fuel) is a redox reaction!

Redox Reactions

Reduction Potentials

We just discussed in the last section how a redox reaction is really two reactions that happen together, an oxidation and a reduction. It makes sense that they occur together if we think about how oxidations lose electrons while reductions gain electrons (remember OIL RIG). The two reactions must be coupled because you can’t have an electron donor without an electron recipient. Because of this, these two reactions (oxidation and reduction) are considered to be two halves of a whole. In other words, a redox reaction consists of an oxidation half reaction and a reduction half reaction.

Let’s return to the OIL RIG definition of redox reactions for a moment. It can tell us which compounds are oxidized or reduced, depending on whether electrons are gained or lost. But which atoms gain electrons, and which ones lose them?

Remember that electrons are tiny charged particles that orbit the nucleus of an atom at dizzying speeds. Depending on how many electrons are orbiting the nucleus, and in which orbitals (at what distance from the nucleus), some atoms hold on to some of their electrons very loosely, and have a higher tendency to give them away. Other atoms hold onto their electrons very tightly, and are in fact able to attract nearby electrons into their orbitals. The potential, or tendency, to gain electrons, is the reduction potential. The reduction potential is the tendency of a compound to be reduced.

The reduction potential is experimentally determined for the oxidation and reduction half reactions, and is called E₀. To remember the relationship of E₀ and oxidation and reduction, consider the reduction potential as how much a compound wants to get reduced (it has a lot of potential for reduction). Remember, reduction is a gain of electrons (OIL RIG), so the larger (more positive) the E₀, the more likely the compound in the half reaction is to gain electrons (be reduced).

Think of it as being like a large magnet. A large magnet attracts nails more strongly, and holds them more tightly, than a small magnet. Similarly, a compound with a large (positive) reduction potential attracts electrons more strongly than a small (or even a negative) reduction potential.

The E₀ values can be looked up in tables, and are always given for reduction half–reactions (reactions gaining electrons). For example:

S + 2H⁺ + e^– =  H₂S E₀ = -0.23 V

We know that a redox reaction consists of both a reduction and an oxidation. Shouldn’t there be such a thing as an “oxidation potential,” too? As it turns out, you can tell how likely a compound is to be oxidized from the reduction potential. Since, as mentioned above, a high (large positive) reduction potential means a high tendency to be reduced, it must follow that a lower reduction potential must mean that a compound is less likely to be reduced.

If a compound is less likely to be reduced, it is more likely to be oxidized. Thus, in a redox reaction, when two half-reactions are coupled together, the half-reaction (with the higher E₀) will be the reduction reaction, while the half-reaction with a smaller (or more negative) E₀ will then be forced backwards and act as the oxidation reaction. A knowledge of the half-reaction reduction potentials can therefore tell you whether compounds are more likely to be oxidized or reduced. It’s like a tug-of-war. The strongest team always pulls the weaker team into the mud pit. In the same way, the compound with the larger reduction potential pulls the electrons over to its side.

Finally, the tendency of the whole reaction to proceed can be calculated from reduction potentials of half reactions, which have been determined experimentally, using the formula:

DE₀ = (E₀ from reaction where e- are gained) – (E₀ from reaction where e- are released)

A positive DE₀ tells you that the reaction will proceed in the direction it is written.

Let’s do an example to make all these rules seem a little more clear. In a reaction that is needed for ATP generation in our bodies, cytochrome C molecules are oxidized in a redox reaction where oxygen that we breathe is reduced:

4 cytochrome c₂⁺ + 4 H⁺ + O₂  4 cytochrome c₃⁺ + 2 H₂O

The two half-reactions are:

Which of the half reactions is more likely to gain electrons? What is the overall DE₀ of the redox reaction?

Notice how the half reactions are both written as reductions. The positive (larger) value for E₀ means that the half reaction will insist on gaining electrons, and act as the reduction. Reaction (2) has the higher E₀, and is thus more likely to gain electrons. Because reaction (1) is by comparison less likely to gain electrons, reaction (1) will be forced backwards, becoming an oxidation reaction, by reaction (2).

Thus, in order to “add” the two half reactions together to get the whole reaction, we will have to reverse half reaction (1):

cytochrome c₂⁺ = cytochrome c3+ + e- [reverse of (1)]

Now let’s add the two half reactions to get the whole redox reaction. Notice that we will have to multiply the above reaction [reverse of (1)] by 4, and reaction (2) by 2, to get the final reaction. However, the E₀ values are not multiplied! This is because the DE₀ value is independent of the numbers of electrons that are in the half reactions.

[reverse of (1)]	4 cytochrome c2+  4 cytochrome c3+ + 4 e–	(e- released)
(2)	O2 + 4 H+ + 4 e–  2 H2O	(e- gained)
(SUM)	4 cytochrome c2+ + 4 H+ + O2  4 cytochrome c3+ + 2 H2O

DE₀ = (E₀ from reaction where e- are gained) – (E₀ from reaction where e- are released)

DE₀ = (0.815 V) – (0.235 V)

DE₀ = 0.580 V

Energy on Tap

Redox reactions are powerful things. That’s not just a figure of speech. Redox reactions really are useful for generating power. Remember how we defined redox reactions as being comprised of the transfer of electrons? That’s exactly what electricity is, the movement of electrons. You may have noticed that the reduction potential (E₀) of half reactions is in Volts, which we use as a measure of electricity?

Here’s an example. Let’s say you buy some alkaline batteries to power your electric toothbrush. When you hit the “on” switch, how do batteries generate the electricity to run that gadget?

 

The battery has zinc powder at one end, and manganese dioxide at the other end. The two ends are connected by a potassium hydroxide (KOH) solution that provides OH^– ions.

The electrons flow out of the battery, and this electric current powers those motorized bristles, letting you get those pearly whites sparkling clean. Moving along the circuit, the electrons then re-enter the battery:

The OH^– ions are regenerated, and are free to flow through the KOH solution to the Zn to generate more electrons.

Your body uses redox reactions to create energy, too, although in a different way. The majority of the energy generated in our bodies comes from aerobic metabolism, which uses the O₂ that we breathe to “burn” the fuel that we eat. The oxidation of compounds such as carbohydrates in redox reactions makes the generation of ATP, our “energy currency”, possible. There are other energy-producing redox reactions in the body that do not utilize oxygen directly.

For example, let’s say you start to get ready for the upcoming ski season by doing some leg lifts. Under non-strenuous conditions, aerobic metabolism uses O₂ to generate ATP to power the muscle. However, during vigorous exercise where O₂ demand is very high in the muscles, an interesting reaction called homolactic fermentation is responsible for allowing our muscles to keep working hard even when oxygen starts to run low (such as by the time you get to your 100th leg lift!). The overall reaction is:

Pyruvate + NADH + H⁺  lactate + NAD⁺

This is a redox reaction that consists of two half reactions:

The NAD⁺ that is generated is then used in other metabolic reactions to generate more ATP. The lactate (lactic acid) generated by this reaction is believed to be responsible for the “burn” that you feel in muscles that you worked too hard.

 

§      When black silver sulfide forms on your favorite piece of silver jewelry, a redox reaction has occurred.

§      When rust forms on your Ford Mustang, a redox reaction has occurred.

§      When you fuel your iPod using batteries, a redox reaction has occurred.

§      When gasoline combusts with oxygen and fuels your car, a redox reaction has occurred.

§      When the natural gas you use to heat your home in the cold Manitoba winters reacts with oxygen gas, a redox reaction has occurred.

§      The process of photosynthesis in plants that produces the air we breathe is a redox reaction.

More often than not, redox reactions can be the culprit or the cure for many of the chemical reactions you observe in everyday life.

There are several types of chemical reactions:  Synthesis, decomposition, single replacement, double-replacement, combustion and acid-base.  Interestingly, most of these reactions (with the exception of double replacement and acid-base) are oxidation-reduction (redox) reactions.  Redox reactions are classified by having both an oxidation reaction and a reduction reaction, and hence, an oxidizing agent and a reducing agent.  This makes sense since as one reactant is losing electrons (being oxidized), the other is gaining electrons (being reduced) Oxidatioumbers can be helpful in determining whether a reaction is redox or non-redox.  When a change in oxidatioumber occurs in a reaction, with both an increase iumber and a decrease iumber, then the reaction is classified as redox.  If this does not occur, then the reaction is non-redox.

 

The types of chemical reactions are summarized below:

Types of Chemical Reactions:

1.                     Synthesis (composition)

Example:  Tarnishing of silver to form black silver sulfide

     2Ag       +        S         à        Ag₂S

2.                     Decomposition

Example:  Electrolysis of water

    2H₂O        à         2H₂        +        O₂

3.                     Single Replacement

Example:  Formation of zinc carbonate (used in suntan lotion) 

  Zn     +     H₂CO₃    à     H₂    +    ZnCO₃

4.                     Double Replacement

Example:  Formation of barium sulfate (used in X-rays)

BaCO₃  +  Na₂SO₄   à   BaSO₄  + Na₂CO₃

5.                     Hydrocarbon Combustion*

Example: Combustion of methane gas

CH₄    +    O₂    à    CO₂    +    H₂O

*Products are ALWAYS CO₂ and H₂O!

6.                     Acid-Base Reactions*

*This reaction is a form of a Double Replacement reaction where an acid reacts with a base (also called neutralization).

Example:  Milk of Magnesia and HCl (neutralization of stomach acid)  

  Mg(OH)₂     +     2HCl     à     MgCl₂     +    2H₂O

 

*Products are ALWAYS a salt (ionic compound) and water!

Redox or Non-redox?

Use oxidation numbers to determine if a reaction is redox or non-redox.  Rules for assigning oxidatioumbers are found at the end of this document.

Redox Reaction:  A chemical reaction (both oxidation and reduction) in which changes in oxidatioumbers occur.

Oxidation Reaction:  any chemical reaction in which an element increases in oxidatioumber or loses electrons.

                       oxidation

                 0                 +1

Example:     2Na   +    Cl₂    à   2NaCl   

      (Na is being oxidized)

Reduction Reaction:  any chemical reaction in which an element decreases in oxidatioumber or gains electrons.

                                  reduction

                         0              -1

Example:     2Na   +    Cl₂    à   2NaCl   

                         (Cl is being reduced)

NOTE: oxidation cannot occur without reduction!

A helpful mnemonic: **LEO   says   GER**

(Loss of Electrons; Oxidation) (Gain of Electrons; Reduction)

 

Oxidizing Agent:  the substance that causes the oxidation of another element; contains the substance being reduced.

Reducing Agent:  the substance that causes the reduction of another element; contains the substance being oxidized.

  If oxidized   à  a reducing agent

  If reduced    à  an oxidizing agent

                           0                0                   +1 -1

Example:     2Al    +     3Cl₂   à    2AlCl₃

Al is oxidized; Al is the reducing agent

Cl is reduced; Cl is the oxidizing agent      

              0       +2  -1          +2 -1              0

Example:     Zn   +  CuCl₂  à  ZnCl₂  +  Cu

Zn is oxidized; Zn is the reducing agent

Cu is reduced; CuCl₂ is the oxidizing agent

NOTE:  If there is no change in oxidatioumbers, it is not a redox reaction!

Objectives:

§      To learn to identify redox reactions.

§      To study practical applications of redox reactions.

§      To learn to identify substances oxidized and substances reduced, as well as oxidizing agents and reducing agents in redox reactions.

In this lab you will observe a series of demonstrations and perform some experiments. Describe briefly your observations for every reaction and write balanced equations.  For each equation do the complete redox analysis, that is:

1. Assign oxidatioumbers for all elements in the equation;

  2. Identify the element oxidized and the element reduced;

  3. Identify the oxidizing agent and the reducing agent.

 

Materials:

1.                     None required

2.                     Piece of zinc metal, test tube, 6.0 mol/L HCl (3 mL), splint, matches.

3.                     Fume hood, piece of copper metal, 100 mL beaker, 16.0 mol/L HNO₃ (5 mL), distilled H₂O.

4.                     Piece of zinc, 100 mL beaker, 0.10 mol/L copper(II)sulfate (3 mL).

5.                     Piece of copper, 100 mL beaker, 0.10 mol/L silver nitrate (3 mL).

6.                     Tarnished silver (ex. Jewelry), 400 mL beaker, baking soda (NaHCO₃), salt (NaCl), aluminum foil, hot plate, distilled water.

7.                     0.10 mol/L silver nitrate (3 mL), test tube, fume hood, concentrated ammonium hydroxide solution, eyedropper, 0.80 mol/L potassium hydroxide (1.5 mL), 0.50 mol/L glucose (dextrose) solution (1 mL).  

 

Demonstrations/Experiments:

1.                     Burning of natural gas, CH₄, in the presence of oxygen. (DEMO)

You have observed the burning of natural gas, oumerous occasions.  Now write a balanced equation for the combustion of methane, CH₄ and do the redox analysis of the equation.  You can skip the observations for this one.

To practice writing and balancing equations for the combustion of hydrocarbons, write balanced equations for the combustion of propane, C₃H₈, and of gasoline, C₈H₁₈.  You don’t need to do the redox analysis for these equations.

A car engine: site of combustion

The following site shows a great animation of combustion in a car engine:

http://www.youtube.com/watch?v=OXd1PlGur8M&feature=related

 

2.                     Reaction of hydrochloric acid and zinc.

Place a piece of zinc metal into a test tube with 3 mL of a 6.0 mol/L HCl solution.  The produced gas will be collected in an inverted test tube and ignited.  Observe and record the changes.  Write two balanced equations and do the redox analysis for both:

a) Reaction of zinc with acid:

b) Reaction of produced hydrogen gas with oxygen gas:

Animation of Metals reacting with HCl (see Activity 4)

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/redox/home.html

 

3.                     Reaction of nitric acid and copper.  (DEMO)

UNDER THE FUME HOOD put a piece of copper metal into a 100 mL beaker with 5 mL of a 16.0 mol/L HNO₃ solution.  Observe and record the changes.  When the reaction has stopped, fill the beaker with distilled water and leave it under the fume hood.   DO NOT TAKE THE BEAKER OUT FROM THE FUME HOOD!  Write balanced equation for this reaction and do the redox analysis.

Nitric acid & copper penny on video:

http://www.metacafe.com/watch/650550/nitric_acid_acts_upon_a_copper_penny_experiment/

And a more humorous version of the copper/nitric acid experiment….

http://www.stumbleupon.com/demo/?review=1#url=http://www.eagan.k12.mn.us/fletcher/acn/flash/nitric_copper.html

Activity Series Experiments:  (#4-6)

Activity Series Animations (see Activities 1-3)

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/redox/home.html

 

4.                     Metal activity series: zinc and copper.  (EXPT.)

Put a piece of metallic zinc into a 100 mL beaker and cover it with 3 mL of a 0.1 mol/L copper(II)sulfate solution.  Observe and record the changes.  Write a balanced equation for this reaction and do the redox analysis.

Which metal is more active: zinc or copper?

 

5.                     Metal activity series: copper and silver.  (EXPT.)

Put a piece of metallic copper into a 100 mL beaker and cover it with 3 mL of a 0.1 mol/L silver nitrate solution.  Observe and record the changes.  Write balanced equation for this reaction and do the redox analysis.

Which metal is more active: copper or silver?

 

6.                     Metal activity series: aluminum and silver – removal of tarnish. (DEMO)

This method can be used to clean any object that is solid silver (it should not have any gems attached).  In a (400 mL) beaker, prepare solution containing about 1% of baking soda, NaHCO₃, and about 1% of table salt, NaCl.  The amount of solution and size of the beaker depends on the size of the silverware.  Put a piece of aluminum foil at the bottom of the beaker.  The tarnished silver should be in contact with the aluminum foil and completely submersed.  Heat the solution to gentle boil and continue to boil for several minutes.  Remove the silver from the beaker, rinse it and dry it with a paper towel.  Write a balanced equation and do the redox analysis.

Which metal is more active: aluminum or silver?

 

7.                     Reduction of silver – the silver mirror reaction (DEMO)

To a clean test tube add 3 mL of a 0.1 mol/L silver nitrate solution.   To the same test tube, under the fume hood, add drop-wise, mixing concentrated ammonium hydroxide until the gray silver hydroxide precipitate forms (about 1-2 drops only!).  Continue adding concentrated ammonium hydroxide while swirling until the solution JUST becomes clear (maybe only another drop!).   Add 1.5 mL of a 0.80 mol/L potassium hydroxide solution to the mixture.   If a precipitate forms, add more ammonium hydroxide while swirling until the solution JUST becomes clear again (about 3 drops!).   Then add 1 mL of a 0.50 mol/L glucose (dextrose) solution.   Immediately mix it and stopper it!  Keep gently swirling the test tube to cover the whole surface of the tube with the liquid.  Observe what happens.  Be patient, the reaction takes several minutes.  In this reaction, silver ion, Ag⁺, is reduced to metallic silver (the silver cation gains an electron).  Glucose is oxidized.  Silver is the oxidizing agent and glucose is the reducing agent.  The exact reaction is fairly complicated.

This process does not require any electricity and is called “electroless plating.”

Dextrose, a reducing sugar, is used to reduce silver ions in Tollens’ reagent to silver metal, which is then deposited on the inside of the test tube. The reactions involved in this activity are summarized below.

 

The final result is shown below:

Summary:

Several redox reactions were carried out in this laboratory session, all of which have practical applications.  Combustion of natural gas, methane, CH₄, is an example of common reactions in which organic compounds, usually hydrocarbons, are oxidized to carbon dioxide and water in a highly exothermic reaction which provides useful energy.  Gasoline is also a mixture of hydrocarbons, and propane gas, C₃H₈ is a hydrocarbon.  Their combustion is a source of energy that moves our cars and delivers heat to our homes.

Metals differ in their ability to be oxidized.  Some metals, e.g., sodium or potassium are very easily oxidized.  They are considered to be very active (reactive).  They are also strong reducing agents.  Other metals, e.g., gold or platinum, are very difficult to oxidize.  They are inactive and they are poor reducing agents.  These differences in reactivity of metals enable production of electric current in batteries. 

Remember that although redox reactions are common and plentiful, not all chemical reactions are redox reactions.

 

All redox reactions involve complete or partial transfer of electrons from one atom to another. 

In this redox reaction between sodium and iodine:

2Na + I₂ –>2NaI

electrons are completely transferred from sodium to iodine resulting in cation Na⁺ and anion I^–.

In the redox reaction between hydrogen and oxygen:

2H₂ + O₂ –>2H₂O

electrons are partially transferred from hydrogen to oxygen.  Oxygen is a more electronegative element than hydrogen.  The electron pair in the covalent bond is shifted toward oxygen resulting in a partial negative charge on oxygen and partial positive charge on hydrogen.  Both reactions above are examples of oxidation-reduction reactions.  The term oxidation refers to the total or partial loss of electrons by one element, and reduction refers to the total or partial gain of electrons by another element.  Oxidation and reduction always occur together (“someone’s gain is always someone else’s loss”).

The electron-accepting substance is called the oxidizing agent because it helps the other element to be oxidized.  The substance that supplies electrons is called the reducing agent because it helps the other element to be reduced.  In other words, the substance oxidized is a reducing agent, and the substance reduced in an oxidizing agent.  In order to determine oxidation or reduction, it is helpful to assign oxidatioumbers to all atoms in the reactants and products.  The rules for assigning oxidatioumbers follow after the optional assignment.

Optional Assignment:

Predict the products of the following chemical reactions and then balance the reactions.  Determine which of the reactions are redox and which are non-redox by assigning oxidatioumbers to each atom.  

1.        Ag     +     S     à                                               (Synthesis)

2.        Fe₂O₃     à                                                         (Decomposition)

3.        Ca     +     O₂     à                                                        (Synthesis)

4.        Al     +     Fe₂O₃      à                                                  (S.R.)

5.        CuF₂     +     H₂SO₄     à                                                       (D.R.)

6.        Mg     +     HCl     à                                                    (S.R.)

7.        C₄H₁₀     +     O₂     à                                                   (Combustion)

8.        LiOH     +     HCl     à                                                (Acid-Base)

9.        Pb(OH)₄     +     KI     à                                                        (D.R.)

10.      C₆H₆     +     O₂     à                                                    (Combustion)

Rules for Assigning Oxidation Numbers

Oxidation Number:  The apparent charge assigned to an atom of an element.  This is the charge an atom would have if the electron pairs in the bond belonged to the more electronegative atom.

Rules for Assigning Oxidation Numbers:

1.                     The sum of the oxidatioumbers of the elements in any neutral atom or molecule is zero.

2.                     The sum of the oxidatioumbers of the elements in any ion is equal to the charge on the ion.

3.                     Fluorine has an oxidatioumber of –1 in all compounds and ions.

4.                     The oxidatioumber assigned to an alkali metal in a compound or ion is always +1.

5.                     The oxidatioumber assigned to an alkaline earth metal in a compound or ion is always +2.

6.                     In most compounds, aluminum is assigned an oxidatioumber of +3, silver is assigned +1 and zinc is assigned +2.

7.                     In most compounds, hydrogen is assigned an oxidatioumber of +1.  Important exceptions are compounds of hydrogen with alkali metals or alkaline earth metals where hydrogen is assigned a charge of –1.

8.                     In most compounds, oxygen is assigned an oxidatioumber of –2.  Important exceptions are peroxides where oxygen is assigned an oxidation number of –1.

9.                     In many compounds, a halogen as a halide is assigned an oxidatioumber of –1.  Exceptions occur when halogens combine with an element more electronegative than itself.  In these cases, the more electronegative element is assigned the negative oxidatioumber.  For electronegativities:  F > O > Cl > others

10.                Sulfur as a sulfide is assigned an oxidation number of –2.  Nitrogen as a nitride is assigned an oxidation number of –3.

11.                When a compound contains two polyatomic ions, it is sometimes necessary to determine the oxidatioumbers of the elements by splitting the compound into its individual ions.

12.                If the above rules do not allow the oxidatioumber to be assigned unambiguously, assume that the oxidation number is the same as another member of the same family.

 

Review of basic principles from Chemistry

Oxidation-Reduction reactions, better known by the short-form redox reactions, involve the transfer of electrons between atoms and molecules during the reaction. They represent one of the major classes of chemical reactions. We can detect redox reactions by monitoring the oxidation states of the atoms in the reactants and products. Any change in these formal oxidation states during the reactions means that a redox reaction has taken place. Oxidation states were introduced in Chem2000, and you should go through as many exercises on assigning oxidation states as possible. You may wish to review the section in your Chem2000 textbook.

Definition of Oxidation States

For reaction involving the transformation of a neutral atom into an ion, e.g., the oxidation of Na to Na+, the identification of the electron transfer (and the assignment of oxidation states) is obvious. The assignment of oxidation states in covalent compounds is far more complex. In order to see which atom in a compound looses/gains electrons, we formally (on paper) decompose a covalent compound into monatomic ions. This is done by formal heterolytic cleavage of bonds according to the electronegativity difference. For example: the H-Cl bond is formally broken by assigning the bonding electron pair to the more electronegative bonding partner (Cl). This gives the fragments: H+ and Cl^–. The charges of these formal fragments are the oxidation states of the corresponding atoms in the covalent compound. In the case of element-element bonds, homonuclear bonds, the bonds are cleaved homolytically (one electron of a single bond (two electrons of a double bond) goes to each of the bonding partners. Oxidation states are usually denoted by Romaumerals.

Example: H₂O₂

1. step: Lewis structure            2.step: Formal bond cleavage

3. step: Charges of fragments  4. step: Indicate oxidation states in Lewis structure

Guidelines for Determining Oxidation Numbers

Oxidation-Reduction Reactions

We find examples of oxidation-reduction or redox reactions almost every time we analyze the reactions used as sources of either heat or work. Wheatural gas burns, for example, an oxidation-reduction reaction occurs that releases more than 800 kJ/mol of energy.

CH₄(g) + 2 O₂(g) = CO₂(g) + 2 H₂O(g)

Within our bodies, a sequence of oxidation-reduction reactions are used to burn sugars, such as glucose (C₆H₁₂O₆) and the fatty acids in the fats we eat.

We don’t have to restrict ourselves to reactions that can be used as a source of energy, however, to find examples of oxidation-reduction reactions. Silver metal, for example, is oxidized when it comes in contact with trace quantities of H₂S or SO₂ in the atmosphere or foods, such as eggs, that are rich in sulfur compounds.

4 Ag(s) + 2 H₂S(g) + O₂(g) = 2 Ag₂S(s) + 2 H₂O(g)

Fortunately, the film of Ag₂S that collects on the metal surface forms a protective coating that slows down further oxidation of the silver metal.

The tarnishing of silver is just one example of a broad class of oxidation-reduction reactions that fall under the general heading ofcorrosion. Another example is the series of reactions that occur when iron or steel rusts. When heated, iron reacts with oxygen to form a mixture of iron(II) and iron(III) oxides.

Molten iron even reacts with water to form an aqueous solution of Fe²⁺ ions and H₂ gas.

Fe(l) + 2 H₂O(l) = Fe²⁺(aq) + 2 OH^–(aq) + H₂(g)

At room temperature, however, all three of these reactions are so slow they can be ignored.

Iron only corrodes at room temperature in the presence of both oxygen and water. In the course of this reaction, the iron is oxidized to give a hydrated form of iron(II) oxide.

2 Fe(s) + O₂(aq) + 2 H₂O(l) = 2 FeO H₂O(s)

Because this compound has the same empirical formula as Fe(OH)₂, it is often mistakenly called iron(II), or ferrous, hydroxide. The FeO H₂O formed in this reaction is further oxidized by O₂ dissolved in water to give a hydrated form of iron(III), or ferric, oxide.

4 FeO H₂O(s) + O₂(aq) + 2 H₂O(l) = 2 Fe₂O₃ 3 H₂O(s)

To further complicate matters, FeO H₂O formed at the metal surface combines with Fe₂O₃ 3 H₂O to give a hydrated form of magnetic iron oxide (Fe₃O₄).

FeO H₂O(s) + Fe₂O₃ = 3 H₂O(s) Fe₃O₄ n H₂O(s)

Because these reactions only occur in the presence of both water and oxygen, cars tend to rust where water collects. Furthermore, because the simplest way of preventing iron from rusting is to coat the metal so that it doesn’t come in contact with water, cars were originally painted for only one reason  to slow down the formation of rust.

Redox (reduction-oxidation) reactions include all chemical reactions in which atoms have their oxidation state changed. This can be either a simple redox process, such as the oxidation of carbon to yield carbon dioxide (CO₂) or the reduction of carbon by hydrogen to yield methane (CH₄), or a complex process such as the oxidation of glucose (C₆H₁₂O₆) in the human body through a series of complex electron transfer processes.

Fundamentally, redox reactions are a family of reactions that are concerned with the transfer of electrons between species. The term comes from the two concepts of reduction and oxidation. It can be explained in simple terms:

·                     Oxidation is the loss of electrons or an increase in oxidation state by a molecule, atom, or ion.

·                     Reduction is the gain of electrons or a decrease in oxidation state by a molecule, atom, or ion.

Although oxidation reactions are commonly associated with the formation of oxides from oxygen molecules, these are only specific examples of a more general concept of reactions involving electron transfer.

Redox reactions, or oxidation-reduction reactions, have a number of similarities to acid–base reactions. Like acid–base reactions, redox reactions are a matched set, that is, there cannot be an oxidation reaction without a reduction reaction happening simultaneously. The oxidation alone and the reduction alone are each called a half-reaction, because two half-reactions always occur together to form a whole reaction. When writing half-reactions, the gained or lost electrons are typically included explicitly in order that the half-reaction be balanced with respect to electric charge.

Though sufficient for many purposes, these descriptions are not precisely correct. Oxidation and reduction properly refer to a change in oxidation state — the actual transfer of electrons may never occur. Thus, oxidation is better defined as an increase in oxidation state, and reduction as a decrease in oxidation state. In practice, the transfer of electrons will always cause a change in oxidation state, but there are many reactions that are classed as “redox” even though no electron transfer occurs (such as those involving covalent bonds).

Redox reactions — reactions in which there’s a simultaneous transfer of electrons from one chemical species to another — are really composed of two different reactions: oxidation (a loss of electrons) and reduction (a gain of electrons).

The electrons that are lost in the oxidation reaction are the same electrons that are gained in the reduction reaction. These two reactions are commonly called half-reactions; the overall reaction is called a redox (reduction/oxidation) reaction.

Oxidation

There are three definitions you can use for oxidation:

·           The loss of electrons

·           The gain of oxygen

·           The loss of hydrogen

Loss of electrons

One way to define oxidation is with the reaction in which a chemical substance loses electrons in going from reactant to product. For example, when sodium metal reacts with chlorine gas to form sodium chloride (NaCl), the sodium metal loses an electron, which is then gained by chlorine.

The following equation shows sodium losing the electron:

Na (s) = Na⁺ + e

When it loses the electron, chemists say that the sodium metal has been oxidized to the sodium cation. (A cation is an ion with a positive charge due to the loss of electrons.)

Reactions of this type are quite common in electrochemical reactions, reactions that produce or use electricity.

Gain of oxygen

Sometimes, in certain oxidation reactions, it’s obvious that oxygen has been gained in going from reactant to product. Reactions where the gain of oxygen is more obvious than the gain of electrons include combustion reactions (burning) and the rusting of iron. Here are two examples.

Burning of coal:

C (s) + O₂ (g) = 2 Fe₂O₃

Rusting of iron:

2Fe(s) + 3 O₂ (g) = 2 Fe₂O₃ (s)

In these cases, chemists say that the carbon and the iron metal have been oxidized to carbon dioxide and rust, respectively.

Loss of hydrogen

In other reactions, oxidation can best be seen as the loss of hydrogen. Methyl alcohol (wood alcohol) can be oxidized to formaldehyde:

CH₃OH (l) = CH₂O (l) + H₂ (g)

In going from methanol to formaldehyde, the compound went from having four hydrogen atoms to having two hydrogen atoms.

Reduction

Like oxidation, there are three definitions you can use to describe reduction:

·  The gain of electrons

·  The loss of oxygen

·  The gain of hydrogen

Gain of electrons

Reduction is often seen as the gain of electrons. In the process of electroplating silver onto a teapot, for example, the silver cation is reduced to silver metal by the gain of an electron. The following equation shows the silver cation gaining the electron:

Ag+ + e = Ag

When it gains the electron, chemists say that the silver cation has been reduced to silver metal.

Loss of oxygen

In other reactions, it’s easier to see reduction as the loss of oxygen in going from reactant to product. For example, iron ore (primarily rust) is reduced to iron metal in a blast furnace by a reaction with carbon monoxide:

2Fe(s) + 3 CO (g) = 2 Fe  (s) + 3 CO₂ (g)

The iron has lost oxygen, so chemists say that the iron ion has been reduced to iron metal.

Gain of hydrogen

In certain cases, a reduction can also be described as the gain of hydrogen atoms in going from reactant to product. For example, carbon monoxide and hydrogen gas can be reduced to methyl alcohol:

CO(g) + 2 H₂ (g) = CH₃OH (l)

In this reduction process, the CO has gained the hydrogen atoms.

One’s loss is the other’s gain

Neither oxidatioor reduction can take place without the other. When those electrons are lost, something has to gain them.

Consider, for example, the net-ionic equation (the equation showing just the chemical substances that are changed during a reaction) for a reaction with zinc metal and an aqueous copper(II) sulfate solution:

Zn(s) + Cu²⁺  = Zn ²⁺+ Cu

This overall reaction is really composed of two half-reactions, shown below.

Oxidation half-reaction — the loss of electrons:

Zn(s) = Zn ²⁺+ 2e

Reduction half-reaction — the gain of electrons:

Cu2+ 2 e = Cu (s)

Zinc loses two electrons; the copper(II) cation gains those same two electrons. Zn is being oxidized. But without that copper cation (the oxidizing agent) present, nothing will happen. It’s a necessary agent for the oxidation process to proceed. The oxidizing agent accepts the electrons from the chemical species that is being oxidized.

The copper(II) cation is reduced as it gains electrons. The species that furnishes the electrons is called the reducing agent. In this case, the reducing agent is zinc metal.

The oxidizing agent is the species that’s being reduced, and the reducing agent is the species that’s being oxidized. Both the oxidizing and reducing agents are on the left (reactant) side of the redox equation.

Usually, you do not have to go through the formal bond cleavage, as shown above. Simple rules has been developed to calculate oxidation states for most of the known compounds whithout drawing Lewis structures. However, by using these rules, you assume a certain Lewis structure.

1.                      Each atom in a pure element has an oxidatioumber of 0. The oxidatioumber of Cu in metallic copper is 0 and is zero for each atom in I₂ or S₈. [always valid]

2.         For ions consisting of a single atom, the oxidatioumber is equal to the charge on the ion. Elements of periodic Groups 1, 2 and 13 form monatomic ions with a positive charge and oxidatioumber equal to the group number. Aluminum therefore forms Al³+ , and its oxidatioumber is +III. (See Section 3.3.) [always valid]

3.         Fluorine is always -I in compounds with other elements. [always valid]

4.         Cl, Br, and I are always -I in compounds except when combined with more electronegative elements, such as O or F. This means that Cl has an oxidatioumber of -I in NaCl (in which Na is +I, as predicted by the fact that it is an element of Group 1). In the ion ClO^– the Cl atom has an oxidatioumber of +I (and O has an oxidatioumber of -II; see Guideline 5).

5.         The oxidatioumber of H is +I and of O is -II in most compounds. Although this statement applies to many, many compounds, a few important exceptions occur.

•   When H forms a binary compound with a metal, the metal forms a positive ion and H becomes a hydride ion, H^–. Thus, in CaH₂ the oxidation number of Ca is +II (equal to the group number) and that of H is -I. [The deciding factor remains the electronegativity of the bonding partner]

•   Oxygen can have an oxidatioumber of -I in a class of compounds called peroxides, compounds based on the O₂^2- ion. For example, in H₂O₂, hydrogen peroxide, H is assigned its usual oxidatioumber of +I, and so O is -I. [If O is bonded to fluorine you have to use the definition given in 7.1.1]

6.           The algebraic sum of the oxidatioumbers in a neutral compound must be zero; in a polyatomic ion, the sum must be equal to the ion charge. Examples of this rule are the previous compounds and others found in Example 4.8.

Oxidation and Reduction – a working definition

The following basic definition is used to describe redox processes:

If Y gains one or more electrons, it is reduced, and is the oxidizing agent.

If X loses one or more electrons, it is oxidized, and is the reducing agent.

These are definitions that must be learned cold. They are the “oui” and “non” of electrochemistry. All the other ideas are based on them. The origin of the term oxidation comes from the fact that combination with oxygen (e.g. in combustion) is one common form of an oxidation reaction. The element oxidized loses electrons to the oxygen atom. For example, when magnesium is burnt:

Mg = Mg²⁺ + 2 e^–

while oxygen gains electrons:

1/2O₂ + 2 e^– = O^2–

and the net reaction is:

Mg + O₂ = MgO

Reduction originates in the concept of reducing an ore, usually a metal oxide, to the elemental form, usually the pure metal.

For example when iron is made from iron ore, the iron reaction is:

Fe²⁺ + 2 Fe³+ + 8 e^– = 3 Fe

and the carbon monoxide reaction is:

4 CO = 4 CO²⁺ + 8 e^–

Here the oxide ions are spectators, transferring from the iron to the oxidized carbon. The net reaction therefore becomes:

Fe₃O₄ + 4 CO = 3 Fe + 4 CO₂

Balancing redox equations in solution

These examples illustrate the common technique of separating redox reactions into two complementary redox half­reactions. In the above examples, the Mg/Mg²+ reaction is an oxidation half reaction. In such reactions, electrons are always among the products. The CO/CO²+ reaction is also an oxidation half reaction. The other two are reduction half­reactions, for which electrons will always be reactants. Although simple reaction such as these are easy to balance at sight, and we often don’t stop to think that they are in fact redox reactions, more complicated redox reactions need a rigorous approach for successful balancing. We will concentrate on reactions that take place in solution, and these will usually be either basic or acidic solutions. Since electrochemistry is a branch of thermodynamics, we usually will be dealing with reactions under conditions of standard state, which for solution chemistry                    means 1 M concentrations at 25°C. Thus acidic solutions will have [H3O+] =                     1.00 M, while basic solutions will have [OH ] = 1.00 M. The usual K_w  relationship between hydroxide and hydronium ion will always hold.

The rules for balancing redox reactions are as follows. These are better learned by doing many examples than by actual memorization of the rules!

Step 1. Assign oxidation states and recognize the reaction as an oxidation-reduction.

Step 2. Separate the overall process into half-reactions.

Step 3. Balance each half-reaction for material (first balance oxidized/reduced element, then balance O, then balance H). In acid solution, add H₂O to the side requiring O atoms.

Then add H+ to balance any remaining unbalanced H atoms.

In basic solution, add H₂O to the side requiring O atoms.

Then add H2O to the side requiring H atoms, and one OH^– to the other side. Add H2O/OH^– for every unbalanced H atom.

Step 4. Balance the half-reactions for charge by inserting the correct number of electrons. Consider the number of oxidized/reduced atoms.

Step 5. Multiply the balanced half-reactions by appropriate factors to obtain the same number of electrons in both half­reactions.

Step 6. Add the balanced half-reactions.

Step 7. Simplify the equation by eliminating common reactants and products.

Step 8. Check the final result for material and charge balance.

Types of Redox Reactions:

Combination

Combination reactions are some of the simplest redox reactions and as the name suggests involves the “combining” of elements to form a chemical compound. As usual, oxidation and reduction occur together. General Equation:

A + B → AB

Sample 1.

Equation: H₂ + O₂ → H₂O

Calculation: 0 + 0 → (2)(+1) + (-2) = 0

Explanation: In this equation both H₂ and O₂ are free elements and following Rule 1, their oxidation state is “0.” The product is H₂O, which has a total oxidation state of “0.” According to Rule #6, the O.S. of oxygen is usually -2. So, the O.S. of H₂ must be +1.

Decomposition

General Equation: AB → A + B

Decomposition reactions are the reverse of combination reactions, meaning they are the breakdown of a chemical compound into the individual elements.

Sample 2.

Equation: H₂O → H₂ + O₂

Calculation: (2)(+1) + (-2) = 0 → 0 + 0

Explanation: In this equation the water is “decomposed” into a Hydrogen and Oxygen. Similar to the previous sample the H₂O has a total oxidation state of “0,” thus according to Rule#6 the O.S. of oxygen is usually -2 so the O.S. of H₂ must be +1.

Displacement Reactions

Displacement reactions, also known as replacement reactions, involve compounds and the “replacing” of elements. They occur as single replacement and double replacement reactions.

Single Replacement

General Equation: A + BC → AB + C

A single replacement reaction involves the “replacing” of an element in the reactants with another element in the products.

Sample 3.

Equation: Cl₂ + NaBr → NaCl + Br₂

Calculation: (0) + ((+1) + (-1) = 0) -> ((+1) + (-1) = 0) + 0

Explanation: In this equation Br is replaced with Cl and Cl is reduced, while Br is oxidized.

Double Replacement

General Equation: AB + CD → AD + CB

A double replacement reaction is similar to a double replacement reaction, but involves “replacing” two elements in the reactants, with two in the products.

Sample 4.

Equation: Fe₂O₃ + HCl → FeCl₃ + H₂O

Explanation: In this equation Fe and H trade places and oxygen and chlorine trade places.

Combustion

Combustion reactions always involve oxygen, in the form of O₂ and are almost always exothermic, meaning they produce heat.

General Equation: C_xH_y + O₂ → CO₂ + H₂O

Disproportionation

General Equation: 2A → A’ + A”

In some redox reactions substances can be both oxidized and reduced. These are known as disproportionation reactions, which have some practical significance in everyday life including the reaction of hydrogen peroxide, H₂O₂ poured over a cut. This a decomposition reaction of hydrogen peroxide, which produces oxygen and water. Oxygen is present in all parts of the chemical equation and as a result it is both oxidized and reduced.

Reaction: 2H₂O₂(aq) → 2H₂O(l) + O₂(g)

Explanation: In the reactants H has an O has an O.S. of -1, which changes to -2 for the product, H₂O (reduced) and 0 for the product, O₂ (oxidized).

 

Example: 

Balance the following equations in both acidic and basic environments:

1) H₂(g) + O₂(g)  → H₂O(l)

Acidic Answer: 2H₂(g) + O₂(g) → 2H₂O(l)

Basic Answer: 2H₂(g) + O₂(g) → 2H₂O(l)

2) Cr₂O₇^2-(aq) + C₂H₅OH(l) → Cr³⁺(aq) + CO₂(g)

Acidic Answer: 2Cr₂O₇^2-(aq) + C₂H₅OH(l) + 16H⁺(aq) → 4Cr³⁺(aq) + 2CO₂(g) + 11H₂O(l)

Basic Answer: 2Cr₂O₇^2-(aq) + C₂H₅OH(l) + 5 H₂O(l) → 4Cr³⁺(aq) + 2CO₂(g) + 16OH^–(aq)

3) Fe²⁺(aq) + MnO₄^–(aq) → Fe³⁺(aq) + Mn²⁺(aq)

Acidic Answer: MnO₄^–(aq) + 5Fe²⁺(aq) + 8H⁺(aq) → Mn²⁺(aq) + 5Fe³⁺(aq) + 4H₂O(l)

Basic Answer: MnO₄^–(aq) + 5Fe²⁺(aq) + 4H₂O(l) → Mn²⁺(aq) + 5Fe³⁺(aq) + 8OH^–(aq)

4) Zn(s) + NO^3-(aq) → Zn²⁺(aq) + NO(g)

Acidic Answer: 3Zn(s) + 2NO₃^–(aq) + 8H⁺(aq) → 3Zn²⁺(aq) + 2NO(g) + 4H₂O(l)

Basic Answer: 3Zn(s) + 2NO₃^–(aq) + 4H₂O(l) → 3Zn²⁺(aq) + 2NO(g) + 8OH^–(aq)

5) Al(s) + H₂O(l) + O₂(g) → [Al(OH)₄]^–(aq)

Acidic Answer: 4Al(s) + 3O₂(g) + 10 H₂O(l) → 4[Al(OH)₄]^–(aq) + 4H⁺(aq)

Basic Answer: 4Al(s) + 3O₂(g) + 6H₂O(l) + 4OH^–(aq) → 4[Al(OH)₄]^–(aq)

 

Example: 

Copper metal added to concentrated nitric acid

Cu_(s)  +  HNO_3(aq) —–>  Cu(NO₃)_2(aq)  +  NO_2(g)

a. first divide the equation into half-reactions (notice that the copper is going from a 0 oxidation state to 2+ which is oxidation, and some of the nitrogen is being reduced from 5+ in the nitrate ion to 4+ in the nitrogen dioxide).

Oxidation:  Cu_(s) —–>  Cu²⁺_(aq)  +  2e^–

Reduction:  NO₃^1-_(aq)  +  e^– —–>  NO_2(g)

b. balance each half-reaction with respect to atoms first, then with respect to electrons.

Oxidation:  Cu_(s) —–>  Cu²⁺_(aq)  +  2e^–

Reduction:  2NO₃^1-_(aq)  +  2 e^– —–>  2NO_2(g))

Notice that this time there are oxygen atoms in the reduction step that are not balanced. When this happens, add as many water molecules on the right as are needed to balance the total oxygens on the left. Then add hydrogen ions on the left side of the arrow to balance the number of hydrogen atoms that were introduced with the water molecules.

Oxidation:  Cu_(s) —–>   Cu²⁺_(aq)  +  2e^–

Reduction:  2NO₃^1-_(aq)  +  2 e^–  +  4H¹⁺_(aq) —–>  2NO_2(g))  +  2H₂O_(l)

c. add the two half-reactions together cancelling the electrons which are now equal on each side of the arrow.

Cu_(s)  +  2NO₃^1-_(aq)  +  4H¹⁺_(aq) —–>  Cu²⁺_(aq)  +  2NO_2(g))  +   2H₂O_(l)

Note that this is a Net Ionic Equation for the reaction. The other two nitrate ions that would be with the four hydrogen ions in the nitric acid would remain unchanged and provide the nitrate ions that would form copper(II) nitrate.

 

 

Some examples of balancing redox reaction equations

The following reaction takes place in standard acid solution (1 M H⁺)

MnO₄^–(aq) + HSO₃ (aq) =  Mn²⁺(aq) + SО₄^2– (aq)

We recognize that Mn changes oxidation state from +7 to +2, while S changes from +4 to +6. The two half reactions are:

 

Finally, check that the atoms and charges balance at the right and at the left The following reaction takes place in standard basic solution (1 M OH^–)

Fe(OH)₂(s) + CrO₄^2– (aq) = Fe₂O₃(s) + Cr(OH)₄^– (aq)

We recognize that Fe changes oxidation state from +2 to +3, while Cr changes from +6 to +3. The two half reactions are:

 

Finally, check that the atoms and charges balance at the right and at the left

Spontaneous reactions

Gibbs Free Energy of reaction tells us whether a certain reaction is product favoured or reactant-favoured in the forward direction. The electrochemical analogue to this concept is called the cell potential given by the symbol E°. E° is expressed in the common electrical unit of volts, but is really just the Gibbs free energy in disguise! The exact relationship between cell potential and Gibbs energy is given by the relationship:

AG = — nFE in general, and for standard conditions; AG⁰ = — nFE⁰;

n = number of electrons transferred, F = Faraday’s constant = 96485 C mol^-1

Because of the minus sign in the equation, the cell voltage E has the opposite sign convention to that of AG:

Product-Favoured Reaction: -ve                   ∆G    or      +ve E

Reactant-Favoured Reaction: +ve      ∆G      or     -ve E

Why do we have such a confusing convention? The origin of this discrepancy comes from the way that physicists define current flow: although electrical current is entirely due to the migrations of electrons in a circuit, a forward current flow is defined in standard theories of electricity as the direction of positive charge flow. It is when harmonizing standard electrical conventions with chemical definitions that these apparently contradictory conventions are generated. In electrochemistry, a reaction which is product favoured produces a voltage, and is therefore called a Voltaic cell, in honor of Alessandro Volta. A reaction which is reactant favoured can be driven forward by the application of a greater opposite voltage; such reactions are called electrolytic cells, and the process is named electrolysis.

We now want to remember another aspect of the meaning of AG, which is that it tells for any given reaction the maximum amount of the total energy change that may be harnessed for useful work. The electrochemical equivalent of this idea is the maximum voltage of the electrochemical cell. This maximum potential is obtainable only from a fully charged cell running under zero load. This means that we can calculate the maximum cell voltage from standard free energies for the reaction.

Consider the reaction:

Zn(s) + Cu²⁺(aq) = Zn²+(aq) + Cu(s)

A fG° (kJ mol^-1) 0                                       65.52         -147.0          0

Since A_rG° = E(AfG° (products)} – E(A_fG° (reactants)}, ∆_rG° = (-147.0 kJ/mol) – (65.52 kJ/mol) = -212.5 kJ mol^-1, and we would predict this to be a spontaneous reaction. It is spontaneous, but frankly quite useless unless you need some finely divided metallic copper and have only these reactants at hand. Let us now see if it is indeed possible to harness the useful work from this reaction.

Diagram of an electrochemical cell to harness the Zn/Cu reaction

The electrochemical cell

To do this we must perform exactly the same chemical reaction but using an electrochemical cell. A typical cell-design is shown in the figure below. It divides the redox reaction into to compartments, each of which contains one of the two redox half­reactions. What voltage does our device measure?

What voltage should it measure? Well, A_rG° = – 212.5 kJ mol^-1, and the value of n = 2, as can be seen by breaking the reaction into its constituent electrochemical half-reactions:

Any discrepancy between the actual measured potential of a given cell and this value is usually due to non-standard conditions, primarily that the concentrations are not exactly one Molar.

 

Electrode or Half-cell potentials

Although the reaction only requires a copper(II) salt and metallic zinc, we have built a cell in which both metallic zinc and copper are present, and copper and zinc sulfate. The reason for this is to make the reactions reversible. It turns out that accurate potentials can only be measured for reversible electrochemical cells. Electrochemists have developed a short-hand notation to indicate the complete make-up of a given electrochemical cell. It is based on the balanced redox reaction, but includes both the reactants and the products. For our cell the correct notation would be:

Zn(s) Zn²+(1 M) Cu²+(1 M) Cu(s)

In this notation, the cathode reaction is always put on the right hand side, and the anode reaction at the left. The cathodic process is always reduction, while the anodic process is always oxidation. Thus the anode half-cell will always be written at Chemistry 3830 Lecture Notes the left hand side of the notation. A convenient mnemonic device to remember this convention is “Right Red Cat”, i.e. that the cathode process is a reduction process and is placed at the right of the cell notation. The vertical lines, , in the notation indicate a phase boundary, such as between a solid and a liquid. Double lines, , indicate double boundaries, such as commonly occur when a salt bridge is placed between the two half cells. If the components of a redox reaction co-exist in solution without a phase interface, they are listed together with a comma separating oxidized and reduced forms. This is often the case when the current is introduced into a solution via an inert electrode (usually platinum metal); an example is the oxidation of iron(II) to iron (III) for which the notation could be:

Pt(s) Fe²⁺(i M), Fe³⁺(i M).

The operation of a voltaic cell is summarized in Figure 21.4. It represents a complete electrical circuit. In the wires external to the cell, the current is carried exclusively by a moving stream of electrons. However, within the solution, the current must be carried by migrating ionic species, such as in our example cell Zn²+ and Cu²⁺ ions. Within the salt bridge, if one is used, inert ions such as K+ and Cl^– carry the charges and provide for current flow.

Standard Half-Cell Reduction Potentials

From the construction of our electrochemical demonstration cell, and from the cell notation, it should now be obvious to you that electrochemical half-reactions, which we arbitrarily introduced as a tool to balance redox reactions, have some degree of physical reality – they can after all be built into electrochemical half-cells. It should, for example, be possible to uncouple the copper/zinc cell and construct other voltaic cells from them, for example a copper/lithium or a silver/zinc cell, etc. This is indeed possible. It would also be nice to be able to assign a voltage to each half-cell, but this turns out to be impossible, since the voltage depends on a complete electrical circuit being established. Nonetheless, this idea of a half-cell potential is so attractive, that chemists have figured out a way to do so. We do it by arbitrarily setting one redox half-reaction to zero, and measuring all other cells with respect to this arbitrary zero reference point. The universal reference standard for electrochemistry is the standard hydrogen electrode, or SHE. For example, if the zinc half-cell is combined with the SHE, as shown in the Figure below, a voltage is measured as +0.76 V. We assign this cell voltage entirely to the zinc half cell, since SHE is zero. Alternatively, we combine the SHE with the copper half cell, and get +0.34 V. Then we can establish the potential of the copper zinc cell as

E° = + 0.76 + 0.34 V = +1.10 V.

Cu²⁺ (aq) + 2e  = Cu(s)  

2H₂О (l) + H₂(g)  = 2H_aO+ (aq) + 2e

Zn (s) = Zn²+ (aq) + 2e

2H₃O⁺(aq) + 2e~= H₂(g)+2H₂O(l)

Net reaction: 2 H₃О (l) + H₂(g) + Cu²⁺(aq) = 2H₃O+ (aq) + Cu (s)

Net reaction Zn(s) + 2H₃О (aq) = H₂(g) + Zn²⁺(aq) + 2H₂O (l)

Since for any given combination of two half-cells, we caever be sure which half will act as the anode and which as the cathode, electrochemists have a standardized way of expressing half-cell potentials, and that is always to depict them as reductions. This leads to the compilation of the standard reduction potentials in aqueous solution which are compiled in the Table below.

We can use this table in numerous ways. First of all, we caow construct an electrochemical cell out of any combination of half cell reactions. Thus we can go and obtain potentials for the examples cited above:

 

Constructing cell potentials from standard half-cell potentials

copper/lithium cell

From the table we get the half-reactions:

Cu ²+ (aq) +                                             2 e ^– = Cu(s)        E° = +0.337 V

Li + (aq) + e = Li(s)                                 E° = -3.045 V

We now combine them in such a way as to get a positive overall cell potential. This means we must reverse the lithium equation, making it the anode (where the oxidation will take place.)

Li(s) = Li + (aq) + e ^– E° = +3.045 V

now we have the cell potential for the overall reaction:

Cu ²+ (aq) + 2 Li(s) = Cu(s) + 2 Li + (aq) E°cell = +3.382 V

Note here very carefully, that the voltage was not doubled when the coefficients are doubled. However, n for this reaction = 2. Cell voltages are therefore independent of stoichiometry. The stoichiometric information is stored in the value of n.

silver/zinc cell

From the table we get the half-reactions:

Zn ²+ (aq) + 2 e ^– =  Zn(s)          E° = -0.763 V

Ag + (aq) + e ^– = Ag(s)              E° = +0.80 V

We now combine them in such a way as to get a positive overall cell potential. This means we must reverse the zinc equation, making it the anode (where the oxidation will take place.)

Zn(s) = Zn ²+ (aq) + 2 e ^_               E° = +0.763 V

now we have the cell potential for the overall reaction

2Ag+ (aq) + Zn(s) = 2Ag(s) + Zn²+(aq)           E°cell = +1.563 V

These examples should be sufficient to show how cell potentials are determined for standard reduction potentials. Remember that voltaic cells must have a positive cell potential.

Other uses for the table of electrode potentials

In this course, we will not be concerned much with constructing electrochemical cells. But we do want to discuss the chemical behaviour of the elements. We can use the data in the Table directly in the description of chemical properties.

a)               Spontaneity of reaction

Combine any two standard reduction reactions by reversing one of the couples and adding the resulting potentials together. If the E cell is positive, the reaction as written is spontaneous. If negative, the reverse reaction is spontaneous.

b)              Identifying reducing agents

Any species with a negative Eelectrode will tend to be a reducing agent, and elements with large negative electrode potentials are strong reducing agents.

c)               Identifying oxidizing agents

Any species with a positive Eelectrode will tend to be an oxidizing agent, and elements with large positive electrode potentials are strong oxidizing agents.

Latimer diagrams

The main disadvantage of the Table of Redox Potentials is that it is organized by the voltage of the process, rather than by the chemical species involved. This is great if you want to build a battery, or to use an electrochemical process in an analytical technique, e.g. to monitor the concentration of a given species. But it is not great if you want to understand the complex redox chemistry of an element such as iron, copper or manganese. To address this issue, Latimer devised a very helpful way to display redox potentials for the elements. These are given in an Appendix in the text by Shriver, Atkins and Langford.

We illustrate both the need for, and the method of construction, of Latimer diagrams with the following example: What happens when Fe(s) is reacted with a strong, non-oxidizing, acid (for example, hydrochloric or perchloric acid) ? Now there are two common oxidation states of Fe, 2+ and 3+. We thus need to decide which of these two species is predicted by thermodynamics to form in acid solution?

Reminder: thermodynamic data will predict which reactions ought to occur, but cannot determine whether they happen at an observable rate or not. Most of the redox reactions of inorganic compounds are rapid reactions, but there are many times when thermodynamics predicts more than one possible product, and where the actual product is selected by the rate of reaction. We will say a bit more about this later on. The example dealt with here can be decided unambiguously by thermodynamics.

From the table of reduction potentials, we pick out the data that is available for iron. We then remember that the free energies of chemical reactions, i.e. AG, obey algebra (often called Hess’ Law of Heat Summation). This allows us to combine the two equations to obtain the unknown redox potential relating iron in the 3+ state to the element. The calculations are summarized:

Reaction_________ Potential_________ ∆G = -nFE

Fe²+ + 2e^– = Fe         -0.440 V                   -2 x F x -0.440

Fe³+ + e^– = Fe²+       +0.771 V                  -1 x F x +0.771

Fe³+ + 3e^– = Fe         NOT +0.331V!!!!     = +0.109 F = -3 x F x -0.036

Thus we have obtained the desired potential, the so-called skip potential, for the direct conversion of elemental iron to iron(III):

Fe³⁺ + 3e^– = Fe                E = -0.036 V

and we can answer the original question. When Fe is dissolved in acid, we consider the reverse of reactions 1 and 3. We see that the potential for the formation of Fe(II) is greater than for the formation of Fe(III).

Thus, even though the formation of Fe³⁺ is overall favored, the reaction has an even greater tendency to stop at the Fe(II) stage. We can diagram this effectively by plotting the free energy terms (as multiples of the Faraday constant) as a function of oxidation state. In effect, the reaction will stop in the “well” at Fe²+, because it would cost energy to rise up from the Fe(II) state up to the Fe(III) state.

Latimer diagrams are convenient for the display of all redox potentials relating to a given element, including the skip potentials. Thus the three redox potentials we have considered so far for iron are displayed as follows:

For clarity I have added arrows to the Latimer diagram. Thus the numbers as written are standard reduction potentials, i.e. the sign of the redox potential refers to the reaction as it proceeds from left to right. If you need to consider the reverse reaction, as in our question above, the sign must be reversed. Latimer diagrams do not normally have such arrows, so that you need to remember this convention!

Construction and use of the Latimer diagram for copper

Let’s look at another example of such reasoning. Consider the two half-reactions:

 

 

a)             Oxidation of elemental copper

What happens when Cu(s) is placed in water or acid? The reverse of the reactions leading to Cu⁺ and Cu²⁺ are both negative! Under standard conditions, copper will not be oxidized by acid! Remember that in your lab experiment on the reactions of metals, you had to use HNO₃ to dissolve copper wire. In other words, we needed to add a reagent capable of supplying the energy required to overcome the barrier depicted in the above graph. This depends on the additional redox half­reaction:

2 NO₃^– + 4 H⁺ + 2 e^– = N₂O₄ + 2 H₂O                   E = +0.803 V

The overall reaction is:

Cu + 2 NO₃^– + 4 H⁺ =  Cu²⁺ + N₂O₄ + 2 H₂O E = +0.463

You may also ask why the product of the nitric acid oxidation is Cu(II) rather than Cu(I). We need to work out the potential for an overall reaction to form Cu(I):

Cu + NO₃^– + 4 H⁺ = Cu⁺ +  N₂O₄ + H₂O E = +0.283

However, the overall reaction to Cu+ has E > 0; since this is a less positive EMF, the reaction proceeds directly to Cu²⁺ and does not halt at Cu+. This is of course also clear from the free-energy diagram above.

NOTE: be sure that you can obtain such balanced redox reactions from the three redox half-reactions involved in this question. Past experience has shown that even the best students consistently fail this task!

b)             Disproportionation of Copper (I)

Copper(I) in acid solution has an interesting property. Consider the possible reactions of Cu+: it can be reduced to Cu or oxidized to Cu²⁺. Let us consider both of these processes.

 

NOTE: The combination of these two redox half reactions describe a balanced redox reaction, i.e. effectively the final reaction describes an electrochemical cell. Whenever this is the case, the potentials obtained via the free energy calculation is the same as just adding the redox potentials (with the correct sign!) together.

What we have demonstrated is that Cu⁺ is unstable towards disproportionation, the process in which a compound undergoes an autoredox reaction to produce forms of the element with higher and lower than the original oxidation state. The reverse of this equation, the conversion of Cu and Cu²⁺ to 2 Cu+ is called comproportionation. Thermodynamics tells us that for copper, the disproportionation reaction is product-favored, whereas comproportionation is not product-favored.

Generalized Latimer diagrams

Consider the Latimer diagram for iron in acid solution shown below in the form that it is found in SAL. It is extended by a branch describing iron in the presence of both 1 M [CN^–] and 1 M acid. The potentials change because coordination with CN^– ligands alters the free energy of both the Fe(II) and Fe(III) species.

We remember from our discussion of acid/base chemistry that Fe³⁺ in acid solution is a short-hand representation for the aqua complex, [Fe(OH₂)₆]³⁺, while Fe²⁺ is really [Fe(OH₂)₆]²⁺. Thus the two branches of the diagram actually reflect identical redox processes for iron in the presence of two ligand systems. We caow see at a glance that the hexacyano complex of Fe(III) is more stable than the aqua ion with respect to reduction to Fe(II). Also, the oxidation of Fe(0) to Fe(II) is considerably more favorable in the cyanide/acid mixture than in aqueous acid. A Latimer diagram can thus be extended to include any number of related redox systems. We could just as well construct a branch in which iron was coordinated to ammonia or chloride ions, for example.

In general, anything which alters the free energy of the system will change the redox potentials. The following factors all affect the size of AG :         (1) Concentration

(2)       Temperature

(3)       Other reagents which are not inert

(4)       pH (a special case of (3))

In practice, the most important of these two examples for aqueous element chemistry is pH changes. It has become conventional to construct Latimer diagrams for the two extremes of pH = 0 and pH = 14 (respectively                       1 M acid and 1 M base).

This is shown for the element manganese below:

 

Note that dramatic differences in redox potentials occur for these two sets of conditions. The reason for this is that large numbers of H+ and OH^– ions are usually involved in the redox half reactions. Anything that affects the concentration of these ions will therefore have a dramatic effect on the redox potentials. This is such an important consideration that it has led to the wide-scale use of a graphical presentation of the free energy changes that accompany redox reactions. We now consider such Frost diagrams.

Frost diagrams

Frost diagrams are essentially the same as the graphs of free energy against oxidation state, where ∆G is given in units of nF (i.e. in Volts). They are simply quantitative versions of the graphs we have already been considering. We demonstrate the construction of complex Frost diagrams for the element manganese, which has as complicated a redox chemistry as any element known.

Constructing the Frost diagram for Manganese

A Frost diagram relates the free energy of any given redox state to the energy of the elemental form. We calculate the value nE for the overall redox couple:

X(N)/X(0)

and plot each couple against the oxidation state N. The oxidation state scale must be linear, even if the element does not exist in all possible oxidation states. We therefore need to calculate the value nE for each species in the Latimer diagram. This is most economically done step-wise, as shown in the table below. Thus we start by calculating the two-electron process between Mn²⁺ and Mn, which works out to -2.36 V. For the next step, we recognize that Mn+³ to Mn²⁺ is an additional one-electron step, so that +1.5 V must be added to the previous step to get the value that relates the energy of Mn³⁺ to that of elemental manganese.

 

The results of our calculation are plotted on the graph shown at the right. In addition, the graph also includes the results of a similar type of calculation using the Latimer diagram for basic solution above. This allows us to directly compare the redox behavior of the element in acid and basic solution.

NOTE: you should test your ability to construct Frost diagrams by doing the base calculation and thereby confirming the points on the graph for basic solution.

Reading the Frost diagram for manganese

Most reactions in aqueous solution occur between the extremes of 1 M acid and 1 M base. For neutral solution, a line approximately intermediate between the two lines plotted will be observed. It is rarely necessary to actually calculate the intermediate state in order to explain what happens in a neutral-solution reaction. There are several characteristic features of Frost diagrams that you should be aware of in order to be able to interpret chemical behavior quickly from the graphs. (NOTE: there is unfortunately no universal agreement as to whether the oxidation state in such diagrams are plotted as increasing from left to right, or vice versa. The two text books used for Chemistry 2810, Shriver-Atkins-Langford and Rayner-Canham, differ in this regard. In these notes I will consistently use the most-positive-at-right convention (that used in Shriver-Atkins- Langford), and I will provide alternatives to all the reversed diagrams presented in Rayner-Canham’s book.

a) Identifying oxidizing and reducing agents

Species lying high on the diagram are oxidizing agents towards species on their left, while species high on the diagram act as reducing agents to other oxidizing agents on their right. Another way to visualize this is by considering the lines connecting the higher and lower lying species. If the line has a postitive slope, the higher-lying species is an oxidizing agent. If the line has a negative slope, the higher-lying species is a reducing agent.

Thus for Mn in both acid and base solution, MnO4^– is an oxidizing agent (the line has a positive slope), being reduced to several possible manganese species of lower oxidation state. In acid solution, the remaining forms of manganese down to Mn³⁺ are all potential oxidizing agents, while in base the lowest oxidizing agent is MnO₂. Elemental manganese, (i.e. manganese metal) is a reducing agent (the line has a negative slope), being itself oxidized most readily to Mn²⁺ in acid solution, and Mn₂O₃ in base. In base only, Mn(II) can also act as a reducing agent.

b) Identifying strong and weak agents

Once we have identified potential oxidizing and reducing agents from the slopes of the lines, their relative strength can be determined by the steepness of the slope of the lines. The steeper the slope, the stronger the agent. Thus MnO₄^– is a much more powerful oxidizing agent in acidic solution than in basic solution. On the other hand, metallic manganese is a slightly stronger reducing agent in basic solution.

c) Identifying redox products (unreactive redox states)

Species at the bottom of the graph have low free-energy, thus little tendency to react. The lowest species on the graph are the thermodynamic final product(s) of the redox reactions involving that element. Note that many Frost diagrams display a thermodynamic well. This is the case for manganese, for which the well is Mn²⁺ in acid and Mn₂O₃ (i.e. Mn(III)) in base.

d) Identifying species likely to undergo disproportionation

If a species lies above the line connecting its neighbors, it is thermodynamically unstable towards disproportionation. This has been described as a point lying along a concave line. For example, in basic solution MnO4^3- lies on a point which is above the line connecting MnO₄^2- and MnO₂. This means that the reaction:

2 MnO₄^3- + 2 H₂O =  MnO₄^2- + MnO₂ + 4 OH^–

is predicted to be product-favored.

e) Identifying species likely to undergo comproportionation

Species likely to undergo comproportionation to a third species are located to left and right of a point which lies below the line connecting the two species. Thus in acid solution, a mixture of MnO₂ and Mn are expected to react together to form Mn²⁺. The rate of this reaction may be hindered by the insolubility of both species, but thermodynamically it is favored. Similarly, in base, a mixture of MnO₂ and Mn(OH)₂ should comproportionate to Mn₂O₃. Note an important difference between disproportionation and comproportionation. The former process renders a single species unstable, in that an autoredox reaction can occur at any time. Comproportionation, on the other hand, requires both reactants to be present at the same time, and if one is missing the other species remains stable.

An overview of transition metal redox behavior in the first period

One of the great advantages of the Frost concept is the ability to compare the relative behavior of different species. We are now finally ready to take a more detailed look at the redox behavior of the transition element, which was one of the goals we set for ourselves when we introduced redox tools. Consider the following Frost diagram for all the first-row transition elements in acid solution. Despite the large number of compounds, the voltages of the redox processes are quite different, so that the points do not on the whole obscure each other.

•                 We see that all the metals are potent reducing agents, with the exception of copper, for which the oxidized forms have a higher free energy than the element. The reducing strength of the metals goes down smoothly from calcium to nickel, across the period, with nickel being only a mild reducing agent. The jump to copper is fairly large, but its behavior is consistent with the trend towards weaker reducing power – copper simply has none.

•                 The earlier transition elements favor the +3 oxidation state as the most stable form (bottom of the diagram), while for the latter elements +2 is more stable, sometimes (as for cobalt) considerably more stable.

•                 The elements in the middle of the series – Mn and Fe – have the largest range of accessible oxidation states. But the free energies of the highest oxidation states of these elements are extremely high, and they are all potent oxidizing agents. In fact, only Ti in its highest state (+4) has virtually no oxidizing power (if we ignore the calcium and scandium ions, which immediately form stable noble-gas configurations forms and are extremely stable and totally non-oxidizing.)

Trends in the redox behavior down Group 6

The Frost diagram at the right picks on one of the elements in the above diagram, still in acid solution, and compares its behavior to the remainder of the transition elements in this group. This may be taken as fairly representative of the redox trends down any d-block group. Note that this diagram has not included the chemical form of the species involved. It is

The highest oxidation state for Cr, +6, is strongly oxidizing. For Mo, +6 is mildly oxidizing, but for W it is completely non-oxidizing.

The most stable oxidation states are +3 for Cr, +4 for Mo, and +6 for W. Higher oxidation states become more favorable in the 2^nd and especially the 3^rd transition series.

In the +5 state, the energy of Mo and W species are the same, which leads to very similar behavior for the two elements in this oxidation state.

Kinetics of redox processes

A detailed consideration of the kinetics of redox processes is beyond the scope of this course. However, in order to make sense of redox reactions, particularly those of the non-metals, we will require to take account of a few kinetic principles.

Overpotentials for gas formation

Reactions which produce gases, especially H₂, O₂ and N₂ often have an overpotential associated with them, and tend to be kinetically slow. Overpotentials are simply the extra energy required to overcome activation energies for reactions to occur at appreciable rates. Consider the reaction to initiate the reduction of H₃O+ to H₂:

H₃O⁺ + e^– = 4H₂ + H₂O

Thus the overpotential is simply the voltage needed to sustain a particular rate of electron transfer.

Overpotential (V) for gas evolution at 25 °C

 

For the simple reaction of hydrogen evolution by a metal dissolving in water or acid, the overpotential is ~0.6 V. This explains why Zn, Fe, Ni and Pb do not evolve hydrogen gas when placed in water. Although the E>/₂ for Zn is 0.76 V in 1 M acid, this is reduced considerably when the [H₃O+] is 1 x 10^-7 ieutral solution. [HINT: use the Nernst equation!] Thus we found in the lab that Zn did not react with boiling water, but produced H2 rapidly from 6 M H3O⁺.

ASIDE: The presence of overpotentials is often exploited in the design of electrochemical experiments and industrial cells. Thus redox reactions can be studied at voltages where H₂ or O₂ evolution from water should be occurring, but does not due to the overpotential. This is where the table from Harris becomes useful in finding an electrode material with in this case the highest possible overpotentials. On the other hand, there is a great deal of interest in photoelectrochemistry, in which sunlight is harnessed directly in a suitable electrochemical cell to split water into H₂ and O₂. Such cells require electrodes with the lowest overpotential, and thus they often use platinized Pt (Pt covered with a colloidal deposit of fresh Pt on the surface) as the electrode material.

Outer-sphere electron transfer

If a redox process can occur with no change in the atomic composition of the reacting species, electron transfer is often very fast, and the rates of reaction closely follow the thermodynamic predictions of AG. For example, consider the reaction:

[Fe(phen)₃]³⁺ + [Fe(CN)d^4- = [Fe(phen)₃]²⁺ + [Fe(CNy^3-

This is an example of a reaction which proceeds by an outer-sphere electron transfer, in which the electron is transferred between the two species much as a baton is between two participants in a relay race.

Inner-sphere electron transfer

If a redox reaction requires the transfer of atoms from one reactant to the other, reaction tends to be much slower due to significant activation energies associated with the atom transfer process. The mechanism of such reactions are inner-sphere electron transfers, in which the composition of the coordination sphere of a complex changes during the reaction:

[CoCl(NH₃)5]²⁺ + [Cr(OH₂)₆]²⁺ + 5 H₂O + 5 H+ = [Co(OH₂)₅]²⁺ + [CrCl(OH₂)₅]²⁺ + 5NH₄⁺

The inner-sphere mechanism is common for redox reactions involving oxoanions. For example, the reduction of oxoanions by NO₂^– occurs by attack of N on the oxygen atom of the oxoanion:

NO₂^– + OCl^– = NO₃ + Cl^–

The rate of reduction are found to be:

ClO₄^– < ClOs^–< ClO₂^– < ClO^– and

ClO₄^– < SO₄^2- < HPO₄^2-

Thus the lower the oxidation state of the central atom, the faster the reaction is found to be. Why? Because the O-E bond is strongest for the highest oxidation state, and this bond must be broken for the reaction to proceed. Further evidence in favor of this hypothesis is the effect of size:

ClO₄^– < BrO₄^– < IO₄^–

The strength of the bond is reduced as the central atom size is increased.

Noncomplementary redox reactions

Very slow kinetics are observed when the change in oxidation states in the oxidizing and reducing agent are not the same,

e.g.:

2 Fe³⁺ + Tl+ =  2 Fe²⁺ + Tl³⁺

If this reaction is to proceed stepwise, then one Fe³⁺ reacts with one Tl⁺ to give a very unfavorable Tl²⁺ion. Alternatively the reaction must be fully termolecular, requiring an activated complex containing two Fe³⁺ ions and a Tl⁺ ion. This is also very unlikely, so the reaction tends to be slow.

Pourbaix diagrams

We now wish to discuss the fate of the metallic elements in aqueous solution in somewhat more detail. This ties in with a lot of the work in the lab so far, and has important implications particularly for environmental chemistry.

 Oxidation of the elements by water

The basic reaction we will consider is:

M + H₂O =  M⁺ + 1/2 H₂ + OH^–

This is the reaction involved in oxidation of metals by water, i.e. the process commonly known as rusting.

The tables of standard reduction potentials tells us for which metals this reaction goes in 1 M H₃O+. We must allow for the over potential, ~0.6 V. We must also correct for pH if we want to discuss neutral solutions!

Consider Mg:         E° = -2.37 V is the standard reduction potential in 1 M acid, i.e.

Mg + 2H₃O+ = Mg²⁺ + 2H₂O + H₂  E° = +2.37

Now apply the Nernst equation:. For pH 7 this becomes:

 

So here is a reaction that has enough voltage to overcome the overpotential, so long as a fresh surface is kept exposed to the water such that the oxidation can proceed. This latter condition is what protects Al: a tough coating of Al2O3 protects bulk aluminum from air and water oxidation. This is also why the Mg had to be put in boiling water in the lab for it to react at appreciable rates. A good rule of thumb then is a metal with a reduction potential of —{0.4 + 0.6} = -1V or greater will oxidize in water at appreciable rates in the absence of air. NB: dissolved O₂ will change this picture, of course! We all know that iron rusts in aerated water.)

 Reduction of elements by water

The redox reaction involved in acid solution is:

O₂ + 4 H⁺ + 4 e^– = 2 H₂O               E  = +1.23 V

In basic solution it becomes:

O₂ + 2 H₂O + 4 e^– = 4 OH^–                 E = +0.40 V

Some strong oxidizing agents, e.g. Co³⁺, are reduced by water. The overall reaction in acid is:

4 Co³⁺ + 2 H₂O = 4 Co²⁺ + O₂ + 4 H⁺ E = 0.59 V

This reaction is thus at the overpotential boundary. It becomes fully favored in basic solution, with E = 1.42 V

Stability field of water

We can combine the reduction and oxidation of water with the pH dependence and construct a diagram which represents the stability field for water. This is shown in the diagram at right.

The region boxed in the middle is the normal range found for natural waters, in which water is not oxidized or reduced, and those are the pH ranges found in the various types of natural waters. Specific E/pH zones for different kinds of environments are indicated by the circles drawn into the stability field diagram.

Well aerated natural waters near the surface contain enough dissolved oxygen to get close to the oxidation of water to O₂. Eutrophic lake water contains sufficient dissolved organic matter to approach closely the H⁺ reduction line. Ocean waters are relatively basic, and may be oxidizing if saturated in dioxygen, or reducing if saturated in organic matter (i.e. in stagnant lagoons, etc.).

Fresh waters are considerably more acidic (because of dissolved carbon dioxide), but again they can be oxidizing if saturated in oxygen, or reducing if too much organic matter is consuming all the oxygen. This acidity is greatly enhanced in bogs and organic-laden soils, due to high humic acid content. (Humic acids are complex organic acids occurring in the soil and in bituminous substances formed by the decomposition of dead vegetable matter.)ю

These are strongly reducing conditions, and explain the formation of CH₄ in marshes. Methane was first discovered from this source; Alessandro Volta was one of the first to identify this gas. Ammonia and hydrogen sulfide as well as the inflammable phosphine, PH3, can also emanate from swamps. (There are many rumors of eerie glows emanating in misty swamps; such tales are likely rooted in phosphine being released and burning above the surface of the waters.) If you remember that bogs and marshes release such compounds, in other words very reduced chemical compounds, it is should help you to remember the acidic, reducing character of bog water.

Pourbaix diagrams

The final type of diagram we want to consider for redox chemistry is a type of combination redox/pH predominance diagram developed by the French electrochemistry Pourbaix. The diagrams are usually named after him. These diagrams are closely related to the stability field for water just discussed. Indeed, most Pourbaix diagrams include either the main E/pH lines from the water diagram, or both the main and overpotential lines. What these diagrams add are the relationships between the redox activity and the Bransted acidity of the elements. They are thus related both to Latimer diagrams, and to predominance diagrams for acid/base reactions.

Consider as an example the Pourbaix diagram for iron. Note that some of the lines from the stability field of water are drawn into this diagram, since this diagram applies to aqueous solutions of iron compounds.

•                 The vertical axis plots the standard reduction potential, and the horizontal the pH. Remember what a predominance diagram was: simple vertical boundaries where the most abundant species altered.

•                 The bottom of the diagram refers to reduced species, i.e. Fe(s), or to the type of conditions that lead to reduction.

•                 The top of the diagram to oxidized species and/or oxidizing conditions.

•                 Vertical lines indicate changes in acid base chemistry independent of E°, e.g. Fe³⁺/Fe(OH)₃.

•                 Horizontal lines indicate redox changes unaffected by pH, e.g. Fe/Fe²⁺ below pH 6.

•                 In the more general case, the lines slope, since both E and [H+] or [OH^–] affect the redox process.

To test your ability to read Pourbaix diagrams, see if you can find answers to the following:

(a)    The form of iron which is the strongest oxidizing agent: FeO₄^2- at [H+] < 1 M

(b)   The form of iron which is the strongest reducing agent: Fe(0), elemental iron

(c)                             The predominant form at pH = 7 and E = 0.0 V: Fe(OH)₃ predominates, but close to Fe²⁺/Fe(OH)₂

(d)   E_/° for (acid) reduction of FeO₄^2- to Fe³⁺: 1 M, 0 pH = 0 so E_/° = 2.2 V

(e)    E_/° for reduction of Fe²⁺ to Fe(s):         E_/° = -0.5 V (This MUST be an acid process. Why?)

On the diagram, dashed lines d and e represent respectively the normal and over potential for oxygen evolution according to:

2 H₂O = 4 H+(aq) + O₂ + 4 e^– E°/ = +1.229 V.

The actual E for hydrogen evolution is given by f, while the overpotential is given by line g:

2H+ + 2e^– = H₂ E°_/ = 0.00 V

 

Figure. Predominance area (Pourbaix) diagrams for the /-block elements showing redox chemistry. Adapted from fvl. Pourbaix, Atlas of Electrochemical Equilibria in Aqueous Solutions, National Association of Corrosion Engineers, Houston, 1974. Reprinted with permission.

How elements behave iatural waters

We caow take any of the other Pourbaix diagrams, and compare them to the natural water limits, and predict what forms may exist in various enviromnents. Several lanthanide and actinide element Pourbaix diagrams are shown in the figure at right.

•   What form will Yb take iatural lake water? Answer: Yb(OH)₃

•   Can uranium be solubilized in sea water? Answer: Yes as U0₂²⁺.

•   Would you expect to find cerium metal free in nature?

Answer:

•   Plutonium is highly toxic, as well as being strongly radioactive. What would happen if plutonium was released into a lake or a stream?

Answer:

•   Imagine that plutonium oxides from a nuclear weapons processing centre had been dumped into a small, well aerated lake where the pH = 6-8 and E = 0.0-0.5 V. Over the course of 20 years, the lake converted to a bog, where the pH = 4 and E = 0.1 V. Discuss the enviromnental concerns in the initial stages and the final stages of the lake “storage”, remembering that plutonium would be a serious toxic hazard if it entered the food stream.

•  Answer:

Note that there are some limitations to this approach. First of all, the concentration of the elements iatural waters is often much lower than standard conditions. Furthermore, these data are only valid for pure water plus the element in question. For example, on this basis we would say gold cannot exist in sea water. However, it does, as a chlorocomplex, which is soluble. This is due to a complexation equilibrium. Nevertheless the combined E/pH diagrams provide a very comprehensive insight into the behavior of the elements in aqueous solution, and this is clearly the most important set of conditions relevant to the terrestrial environment.

 

 

Part I. Determination of Standard Half-Cell Potentials

The standard half-cell potentials of Zn/Zn2+, Pb/Pb2+, and Ag/Ag+ will be determined using a copper electrode as a reference. This experimental setup involves cells under near standard conditions (1.0 M, 25°C,1 atm) so the standard potentials are directly measured.

In order to interpret potential results obtained from a voltmeter, it is necessary to understand electrical conventions and how a voltmeter displays the potential it measures. From Ohm’s law we know that V = I.R. Resistance (R) always opposes the flow of current and is always positive. Therefore if the current is positive, the voltage reading also is positive. Positive current from a battery (Galvanic cell) flows from the positive pole at high electrical potential to the negative pole at low electrical potential. Convention dictates that the positive lead (red) be attached to the positive pole and the negative lead (black) be attached to the negative pole to measure the voltage difference of the positive pole relative to the negative pole. If this is done with the batteries found in the laboratory, you will obtain a positive voltage reading. The voltmeter displays the difference of the potential at the positive pole minus the potential at the negative pole. Electrons however flow in the opposite direction as current. So electrons tend to flow from the negative pole of the battery to the positive pole. This means that if the voltage reading on the voltmeter is positive, electrons tend to flow through the black lead into the meter, and out the red lead of the meter. This fact is important to correctly evaluate at which electrode half-cell reduction is occurring.

Because we are using the copper half-cell as the reference, it is connected to the common (negative) terminal of the multimeter. For the copper electrode 0.34 half-cell potential to be taken as the reference value for theses experiments, the over all reaction must be written so that Cu(s) appears as the reactant. Then the voltage measured by the voltmeter will be the potential that tends to push electrons toward the copper electrode. The voltmeter is measuring the difference in the potentials between two half-cells, one of them being copper. The potential displayed is the potential difference of the positive pole minus the negative pole, which would be at a potential of 0.34 V relative to the SHE. If a half-cell of potential X relative to the SHE is attached to the red lead of a voltmeter and a 1.0 M Cu2+/Cu half-cell is attached to the negative terminal of the voltmeter, and if ∆E represents the voltmeter reading, then:

∆E = X – 0.34 or

X = ∆E + 0.34

If the potential, ∆E, is positive then ∆G is negative and the over all reaction occurs in the direction written. Electrons are flowing from the half-cell attached to the negative terminal. If ∆E is negative then ∆G is positive, and electrons are flowing to the half-cell attached to the negative terminal. When reactions that involve other pairs half-cells are measured, the voltmeter will be measuring the potential of the half-cell attached to the positive terminal minus the potential of the half-cell connected to the common (negative) terminal as it tries to push electrons into the negative terminal. A negative value for ∆E simply means that a negative push is a pull. The metal ions originally written as reactants in the half-cell attached to the common (negative) terminal must be written as products to describe the reaction as it occurs spontaneously. 

For example, for the cell Cu/Cu²⁺//Zn²⁺/Zn, the standard reduction potentials are as follows:

Cu²⁺(aq)+ 2e = Cu(s)                                  E°(Cu)                  = + 0.34

Zn²+(aq) + 2e^– =  Zn(s)                               E°(Zn)                  = – 0.76

The symbol // represents a salt bridge, which provides electrical contact between the electrode couples without allowing mixing and, in the ideal case, contributes no liquid junction potential to the cell potential. Because for this experiment we are measuring potentials with the copper electrode at the negative terminal of the voltmeter, we must write the overall reaction with Cu(s) as a reactant:

Zn²+(aq) + Cu(s) =  Zn(s) + Cu²+(aq)           ∆E°(Cu/Zn) = E°(Zn) – E°(Cu) = – 1.10

We can see from this example, that the reaction will occur spontaneously in the direction opposite to the one which it was written. Experimentally we could determine AE°(Cu/Zn) by measuring a potential of -1.10 with a voltmeter. From this measurement we can calculate the half-cell potential for E°(Zn) by substituting for AE in the equation:

X = AE + 0.34

X= -1.10 + 0.34 = – 0.76

Experimental Procedure

1.      Plug the wire leads into the multimeter. The red lead plugs into the VQ hole and the black lead plugs into the COMM hole.

2.                 Attach the alligator clips to the leads if they are not already attached.

3.                 Fill a 50 ml beaker 1/3 full with 1.0 M KNO₃ .

4.                 Obtain a lead, zinc, copper, and silver wire.

5.                 Obtain four soda straws containing an agar/KNO₃ plug at one end.

6.      Label each straw with a marker: Ag, Cu, Pb, Zn.

7.    Using the plastic pipettes supplied fill each straw 4/5 full with the salt   solution of  the appropriate metal.

8.      Place the four straws into the 50 ml beaker containing 1.0 M KNO₃ with the agar tip immersed in the liquid.

9.     Turn on the multimeter to read volts – DC.

10.           Attach the alligator clip to the black multimeter lead to the copper wire. Attach the alligator clip to the red multimeter lead to the silver wire.

11.            Holding the copper wire by the alligator clip, dip the wire into the copper salt solution in the straw labeled Cu.

12.            Holding the silver wire by the alligator clip, dip that wire into the silver salt solution in the straw labeled Ag.

13.            Read and record the potential on the multimeter.

14.            Leaving the copper wire in place remove the silver wire from the solution and release it from the alligator clip.

15.            In sequence pick up the lead then zinc wires with the black alligator clip and measure and record the potential differences with respect to the copper electrode when these wires are dipped into the salt solutions of their metals.

varried concentrations

 

Electrochemical Cell Assembly

From the potential differences obtained with the Cu²⁺/Cu half-cell and the rest of the half-cells, predict and then measure the potential differences for the following reactions:

Pb(s) + 2 Ag⁺(aq) = Pb^[1]+(aq) + 2 Ag(s)                                ∆E°(Pb/Ag) =

Zn(s) + 2 Ag⁺ (aq) = Zn^{2 +} (aq) + 2 Ag(s)                               ∆E°(Zn/Ag) =

Zn(s) + Pb²⁺(aq) =  Zn²+(aq) + Pb(s)                                    ∆E°(Zn/Pb) =

 

Calculations and Results

1. (a) Calculate the standard reduction potentials for each half-cell vs the SHE from the data you collected using the Cu²+/Cu half-cell, given that the E° is 0.34 V for

Cu²+ + 2e~ = Cu(s)

Note that E° for a half-reaction is not dependent on the coefficients, provided, of course, that the reaction is balanced, i.e., E° for

E° for Ag+ + e^– = Ag(s)

is the same as

Ag+ + 2e = 2Ag(s)

 (b) Compare your values with those in Appendix E of Oxtoby. Determine your experimental accuracy for each (deviation of your average value from the true value).

Summarize these results in a table (you may want to use Excel for this).

2. (a) Tabulate the results of the whole cell potential differences that were measured along with their predicted values using the Cu²+/Cu as a reference and the per cent error.

 

Determining Standard State Cell Potentials

A cell’s standard state potential is the potential of the cell under standard state conditions, which is approximated with concentrations of 1 mole per liter (1 M) and pressures of 1 atmosphere at 25^oC.

To calculate the standard cell potential for a reaction

·  Write the oxidation and reduction half-reactions for the cell.

·  Look up the reduction potential, E^o_reduction,  for the reduction half-reaction in a table of reduction potentials

·  Look up the reduction potential for the reverse of the oxidation half-reaction and reverse the sign to obtain the oxidation potential.  For the oxidation half-reaction, E^o_oxidation = – E^o_reduction.

·  Add the potentials of the half-cells to get the overall standard cell potential.

E^o_cell = E^o_reduction + E^o_oxidation

Example:  Find the standard cell potential for an electrochemical cell with the following cell reaction.

Zn(s) + Cu²⁺(aq)  Zn²⁺(aq) + Cu(s)

·  Write the half-reactions for each process.

Zn(s)  Zn²⁺(aq) + 2 e^–

Cu²⁺(aq) + 2 e^–  Cu(s)

·  Look up the standard potentials for the redcution half-reaction.

E^o_{reduction of Cu2+} = + 0.339 V

·  Look up the standard reduction potential for the reverse of the oxidation reaction and change the sign.

E^o_{reduction of Zn2+} = – 0.762 V

E^o_{oxidation of Zn} = – ( – 0.762 V) = + 0.762 V

·  Add the cell potentials together to get the overall standard cell potential.

Top

Determining Non-Standard State Cell Potentials

To determine the cell potential when the conditions are other than standard state (concentrations not 1 molar and/or pressures not 1 atmosphere):

·  Determine the standard state cell potential.

·  Determine the new cell potential resulting from the changed conditions.

o          Determine Q, the reaction quotient.

o          Deternine n, the number of electrons transferred in the reaction “n”.

o          Determine E_cell, the cell potential at the non-standard state conditions using the Nernst equation.

E_cell = E^o_cell – (RT/nF) ln Q

E_cell = cell potential at non-standard state conditions

E^o_cell = standard state cell potential

R = constant (8.31 J/mole K)

T = absolute temperature (Kelvin scale)

F = Faraday’s constant (96,485 C/mole e^–)

n = number of moles of electrons transferred in the balanced equation for the reaction occurring in the cell

Q = reaction quotient for the reaction.  aA + bB  cC + dD,

If the temperature of the cell remains at 25^oC, the equation simplifies to:

E_cell = E^o_cell – (0.0257/n) ln Q

or in terms of log₁₀

E_cell = E^o_cell – (0.0592/n) log Q

Example: Predict the cell potential for the following reaction when the pressure of the oxygen gas is 2.50 atm, the hydrogen ion concentration is 0.10 M, and the bromide ion concentration is 0.25 M.

O₂(g) + 4 H⁺(aq) + 4 Br^–(aq)   2 H₂O(l) + 2 Br₂(l)

·  Calculate the standard cell potential for the reaction, E^o_cell, using the tabled values:

·                     Determine the new cell potential resulting from the changed conditions.

o          Calculate the value for the reaction quotient, Q. (Note: We calculate Q using molar concentrations for solutions and pressures for gases. Water and bromine are both liquids, therefore they are not included in the calculation of Q.)

o          Calculate the number of moles of electrons transferred in the balanced equation, n.

n = 4 moles of electrons

o          Substitute values into the Nernst equation and solve for the non-standard cell potential, E_cell.

E_cell = + 0.152 V – (0.0257/4) ln(1.02 x 10⁶) 
E_cell = 0.063 V

ROLE OXIDATION-REDUCTION REACTIONS ON METABOLISM COMPOUND IN HUMAN BODY

 

Oxidation-reduction reactions are of central importance in organic chemistry and biochemistry. The burning of fuels that provides the energy to maintain our civilization and the metabolism of foods that furnish the energy that keeps us alive both involve redox reactions.

The burning of natural gas is not only a combustion reaction but also a redox reaction. Similar reactions include the burning of gasoline and coal. These are also redox reactions.

All combustion reactions are also redox reactions. A typical combustion reaction is the burning of methane, the principal component of natural gas

CH₄+ 2O₂→ CO₂+ 2H₂O

In respiration, the biochemical process by which the oxygen we inhale in air oxidizes foodstuffs to carbon dioxide and water, redox reactions provide energy to living cells. A typical respiratory reaction is the oxidation of glucose (C₆H₁₂O₆), the simple sugar we encountered in the chapter-opening essay that makes up the diet of yeast:

C₆H₁₂O₆+ 6O₂→ 6CO₂+ 6H₂O

Organic chemists use a variety of redox reactions. For example, potassium dichromate (K₂Cr₂O₇) is a common oxidizing agent that can be used to oxidize alcohols (symbolized by the general formula ROH). The product of the reaction depends on the location of the OH functional group in the alcohol molecule, the relative proportions of alcohol and the dichromate ion, and reaction conditions such as temperature. If the OH group is attached to a terminal carbon atom and the product is distilled off as it forms, the product is an aldehyde, which has a terminal carbonyl group(C=O) and is often written as RCHO. One example is the reaction used by the Breathalyzer to detect ethyl alcohol (C₂H₅OH) in a person’s breath:

3C₂H₅OH + Cr₂O₇²⁻+ 8H⁺ → 3CH₃CHO + 2Cr³⁺+ 7H₂O

If the product acetaldehyde (CH₃CHO) is not removed as it forms, it is further oxidized to acetic acid (CH₃COOH). In this case, the overall reaction is as follows:

3C₂H₅OH + 2Cr₂O₇²⁻+ 16H⁺ → 3CH₃COOH + 4Cr³⁺+ 11H₂O

In this reaction, the chromium atom is reduced from Cr₂O₇²⁻ to Cr³⁺, and the ethanol is oxidized to acetic acid.

When the OH group of the alcohol is bonded to an interior carbon atom, the oxidation of an alcohol will produce a ketone. (The formulas of ketones are often written as RCOR, and the carbon–oxygen bond is a double bond.) The simplest ketone is derived from 2-propanol (CH₃CHOHCH₃). It is the common solvent acetone [(CH₃)₂CO], which is used in varnishes, lacquers, rubber cement, and nail polish remover. Acetone can be formed by the following redox reaction:

3CH₃CHOHCH₃+ Cr₂O₇²⁻+ 8H⁺→ 3(CH₃)₂CO + 2Cr³⁺+ 7H₂O

As we have just seen, aldehydes and ketones can be formed by the oxidation of alcohols. Conversely, aldehydes and ketones can be reduced to alcohols. Reduction of the carbonyl group is important in living organisms. For example, in anaerobic metabolism, in which biochemical processes take place in the absence of oxygen, pyruvic acid (CH₃COCOOH) is reduced to lactic acid (CH₃CHOHCOOH) in the muscles.

CH₃COCOOH → CH₃CHOHCOOH

(Pyruvic acid is both a carboxylic acid and a ketone; only the ketone group is reduced.) The buildup of lactic acid during vigorous exercise is responsible in large part for the fatigue that we experience.

In food chemistry, the substances known as antioxidants are reducing agents. Ascorbic acid (vitamin C; C₆H₈O₆) is thought to retard potentially damaging oxidation of living cells. In the process, it is oxidized to dehydroascorbic acid (C₆H₆O₆). In the stomach, ascorbic acid reduces the nitrite ion (NO₂⁻) to nitric oxide (NO):

C₆H₈O₆+ 2H⁺+ 2NO₂⁻→ C₆H₆O₆+ 2H₂O + 2NO

If this reaction did not occur, nitrite ions from foods would oxidize the iron in hemoglobin, destroying its ability to carry oxygen.

Tocopherol (vitamin E) is also an antioxidant. In the body, vitamin E is thought to act by scavenging harmful by-products of metabolism, such as the highly reactive molecular fragments called free radicals. In foods, vitamin E acts to prevent fats from being oxidized and thus becoming rancid. Vitamin C is also a good antioxidant

Citrus fruits, such as oranges, lemons, and limes, are good sources of vitamin C, which is an antioxidant.

Finally, and of greatest importance, green plants carry out the redox reaction that makes possible almost all life on Earth. They do this through a process called photosynthesis, in which carbon dioxide and water are converted to glucose (C₆H₁₂O₆). The synthesis of glucose requires a variety of proteins called enzymes and a green pigment called chlorophyll that converts sunlight into chemical energy. The overall change that occurs is as follows:

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

Photosynthesis is the fundamental process by which plants use sunlight to convert carbon dioxide and water into glucose and oxygen. Then plants make more complex carbohydrates. It is the ultimate source of all food on Earth, and it is a redox reaction.

 

References:

1. The abstract of the lecture.

2. intranet.tdmu.edu.ua/auth.php

3. Atkins P. W. Physical chemistry / P.W. Atkins. – New York, 1994. – P.299‑307.

4. Cotton F. A. Chemical Applications of Group Theory / F. A. Cotton. ‑ John Wiley & Sons : New York, 1990.

5. Girolami G. S. Synthesis and Technique in Inorganic Chemistry / G. S. Girolami, T. B. Rauchfuss, R. J. Angelici. ‑ University Science Books : Mill Valley, CA, 1999.

6. Russell J. B. General chemistry / J B. Russell. New York.1992. – P. 550‑599.

7. Lawrence D. D. Analytical chemistry / D. D. Lawrence. –New York, 1992. – P. 218–224.

8. http://www.lsbu.ac.uk/water/ionish.html

 

Prepared by PhD Falfushynska H.